旭的小窝

216. Combination Sum III

 
  • 2023年 06月11日

Promble

link: https://leetcode.com/problems/combination-sum-iii/

Find all valid combinations of k numbers that sum up to n such that the following conditions are true:

Only numbers 1 through 9 are used. Each number is used at most once. Return a list of all possible valid combinations. The list must not contain the same combination twice, and the combinations may be returned in any order.

Input: k = 3, n = 7
Output: [[1,2,4]]
Explanation:
1 + 2 + 4 = 7
There are no other valid combinations.

Input: k = 3, n = 9
Output: [[1,2,6],[1,3,5],[2,3,4]]
Explanation:
1 + 2 + 6 = 9
1 + 3 + 5 = 9
2 + 3 + 4 = 9
There are no other valid combinations.

Input: k = 4, n = 1
Output: []
Explanation: There are no valid combinations.
Using 4 different numbers in the range [1,9], the smallest sum we can get is 1+2+3+4 = 10 and since 10 > 1, there are no valid combination.

Approach 1

use itertools libs.

Thought

  1. def res to store the final result.
  2. loop i in combinations(list(range(1,10)), k)
    1. if sum(i) == n: res.append(i)
  3. return res

Code

from itertools import combinations

class Solution:
    def combinationSum3(self, k: int, n: int) -> List[List[int]]:
        res = []

        for i in combinations(list(range(1,10)), k):
            if sum(i) == n:
                res.append(i)
        
        return res

Approach 2

DFS method.

Thought

  1. def temp:list to store the recurrence in def.
  2. def res:list to store the final result.
  3. the dfs start status is: k,n,1,temp,res
  4. def dfs function, var: k,n,start,temp,res
    1. if k == 0 and n == 0: res.append(list(temp)), return
    2. loop i in range(start, 10) # only loop larger start number
    3. judge if i > n: return # pruning
    4. add i to temp
    5. recurrence dfs, self.dfs(k-1, n-i, i+1, temp, res)
    6. temp.pop
  5. return res

Code

class Solution:
    def combinationSum3(self, k: int, n: int) -> List[List[int]]:
        temp = []
        res = []
        self.dfs(k,n,1, temp, res)

        return res 

    def dfs(self, k, n, start, temp, res):
        # example 3
        if k == 0 and n == 0:
            res.append(list(temp))
            return 

        for i in range(start, 10):
            if i > n:
                return # pruning 
            temp.append(i)
            self.dfs(k-1, n-i, i+1, temp, res)
            temp.pop()
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